Problem: $f(x, y, z) = (x^3 - z\sin(y), 2zx, 8x^2)$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\sin(y), x, 0)$ (Choice B) B $(-z\cos(y), 0, 0)$ (Choice C) C $(3x^2, 2z, 16x)$ (Choice D) D $(3x^2 - \sin(y) - z\cos(y), 2z + 2x, 16x)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $x$, we'll treat $y$ and $z$ as if they were constants. Therefore, $f_x = (3x^2, 2z, 16x)$.